How to remove a carriage return (\r\n)
use strict;
use warnings;
{
my $str = "abcd\r\n";
$str =~ s/\r|\n//g;
print "[$str]";
}
{
my $str = "abcd\n";
$str =~ s/\r|\n//g;
print "[$str]";
}
{
my $str = "abcd\r";
$str =~ s/\r|\n//g;
print "[$str]";
}
I blog issues resolved at work
There is also \v which matches both \r and \n.
$g =~ s/\v//g;
Abigail says
Do you know of any difference between the behaviour of \v and \R? They seem to do exactly the same thing, on Perl 5.12.3:
\v:
\R:
See man pages for perlre, perlrecharclass, and perlrebackslash.
\v matches vertical whitespace, which are the characters shown above as of Unicode 6.0: [\x0A-\x0D\x85\x{2028}\x{2029}]
\R matches these and multi-character newline sequences. This means it can't be used inside bracket character matches (e.g. [h\R]). It will match the CRLF sequence (and could match others if more are indicated in later versions of Unicode). As of Unicode 6.0 it is equal to (?>\x0D\x0A?|\v).
Looks like the perlrebackslash man page has a typo in the equivalent regex. It might be tempting to think it would be (?:(?>\x0D\x0A)|\v), but this example shows it isn't implemented that way:
say "v match" if "\x0d\n" =~ /^\v\v$/m;
say "R match" if "\x0d\n" =~ /^\R\R$/m;
matches v but not R. The first \R greedily consumes the sequence.